The Young's modulus of a solid can be expressed as $$ E = \frac{F/A}{\Delta L/L_0} $$ where the force $F$ is exerted perpendicular to surface $A$ resulting in a change in length of $\Delta L$ from the initial length $L_0$.

When the solid is submerged in a fluid (let's say water), an additional force due to the pressure of the water is felt. Does this change the definition of Young's modulus? In other words, if I performed a compression test (i.e. apply a known force and measure displacement) of the same object in air and in water, would I expect to get the same or different results? We can assume the vessel of water is quite large relative to the object.

I am having trouble wrapping my mind around this because one one hand I think the force of water pushing down from above might "pre-stress" the object to some degree, but on the other hand maybe the pressure coming in from the sides eliminates this effect? Furthermore, doe the fact that pressure increases with depth make this challenging?

  • $\begingroup$ Your definition of Young's modulus is only valid for a linear elastic material undergoing small (infinitessimal) strain. A more general definition is $E(\varepsilon) = \text{d}\sigma/\text{d}\varepsilon$, then $\lim_{\varepsilon\to 0} E(\varepsilon)$ agrees with your definition. $\endgroup$
    – Davius
    Jul 15, 2022 at 23:54

4 Answers 4


Young's modulus is generally pressure dependent. (Tabulated values are assumed to be at standard atmospheric pressure unless otherwise stated.)

Here's a simple way to think about it: Stretching a material in one direction tends to increase its volume (by a factor of $1-2\nu$ for isotropic materials, where $\nu$ is Poisson's ratio that never exceeds 1/2 in stable materials). Under tremendous pressures, for example, it wouldn't make sense to obtain this volume increase for free; one has to pull harder to exert the greater work needed to expand the material.

At a more sophisticated level, the reason Young's modulus depends on pressure is that uniaxial stress can be decoupled into a so-called deviatoric stress state and a so-called dilatational stress state.

The former stress state can be thought of as a more sophisticated 3D version of simple shear, although its tensor representation can contain nonshear components. More importantly, deviatoric stresses tend to change the shape of objects, maintaining their volume. For this reason, the relationship between deviatoric stresses and strains isn't really dependent on the surrounding pressure.

The latter stress state is strongly coupled to pressure, though; in fact, hydrostatic pressure is just a negative dilatational stress. Dilational stresses tend to change the size of objects. Materials get stiffer as you squeeze them, and so qualitatively, adding hydrostatic pressure can be expected to alter Young's modulus.

Quantitatively, for isotropic materials, Young's modulus $E$ equals $\frac{9KG}{3K+G}$, where $K$ is the bulk modulus and $G$ is the shear modulus. The former is expected to exhibit positive pressure dependence, and the latter is not expected to be strongly pressure dependent (with zero dependence predicted under the idealization of a deviatoric vs. dilatational dichotomy as discussed above). Thus, we expect a pressure dependence—always positive—of $$\frac{dE}{dP}=\frac{9G^2}{(3K+G)^2}\frac{dK}{dP}.$$ The bulk modulus dependence on pressure $\frac{dK}{dP}$ can be found in the literature for many materials.

To answer your last question, yes, varying depth-dependent pressure does complicate the analysis. Except for simple geometries, you'd likely have to analyze the stress state numerically (using finite element analysis, for instance).

Does this all make sense?


The Young's modulus results from an uniaxial test. For a isotropic elastic material:

$$\sigma_{xx} = (\lambda + 2\mu)\epsilon_{xx} + \lambda \epsilon_{yy} + \lambda \epsilon_{zz}$$ $$\sigma_{yy} = \lambda\epsilon_{xx} + (\lambda + 2\mu) \epsilon_{yy} + \lambda \epsilon_{zz}$$ $$\sigma_{yy} = \lambda\epsilon_{xx} + \lambda\epsilon_{yy} + (\lambda + 2\mu) \epsilon_{zz}$$

Where $\lambda$ and $\mu$ are the so called Lamé parameters. For an uniaxial test in the $x$ direction $\sigma_{yy} = \sigma_{zz} = 0$. Manipulating the equations we can get the relation:

$$\sigma_{xx} = \mu\frac{2\mu + 3\lambda}{\lambda + \mu}\epsilon_{xx}$$

From this expression, the Young's modulus is:

$$E = \mu\frac{2\mu + 3\lambda}{\lambda + \mu}$$

Of course, if $\sigma_{yy}$ and $\sigma_{zz}$ are not zero, the relation between stress and strain in the $x$ direction will be different from $E$.


The definition of Young's Modulus (relationship between tensile stress to tensile strain) doesn't change, but for three dimensional deformation, when the volume is involved, one normally uses the material Bulk modulus.

The Bulk modulus is the ratio of applied stress to volumetric strain. The formula for Bulk modulus $k$ is

$$k=\frac{F/A}{\Delta V/V}$$

where $\Delta V$ is the change in original volume.

For perspective, this site: http://www.matweb.com/search/datasheet_print.aspx?matguid=10b74ebc27344380ab16b1b69f1cffbb has values $E=186$ GPa and $k=148$ GPa for cold rolled steel.

Though both $k$ and $E$ are generally pressure dependent, for most materials the effect of water pressure should be minimal. Water pressure increases about 1 atm (0.1 MPa) for every 10 meters in depth. Even at the bottom of the Mariana Trench, where the pressure is about 109 MPa, the water pressure is a factor of 1000 less than the values of $E$ and $k$ for steel. Even for plastics, such as ABS which for which $k$ is roughly 4 GPa, unless you are at the bottom of the ocean the effect of water pressure should normally be negligible.

Hope this helps.


There would be a small reduction in tensile strain of a rod due to uniform pressure which can be found by calculation from bulk modulus.

However from plasticity failure theory (considering deviatric stress tensor) there would be no change in the ultimate failure by action of hydrostatic pressure on the material.


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